The version in [1]} states
\(\operatorname{Ext}^i_X(E_\star , F_\star \otimes \omega _X(D)) \simeq \operatorname{Ext}^{n-i}_X(F_\star , \widehat{E}_\star )^\vee .\) 
Using [1]}, and taking
\(E_\star = O_X\) , we find
\(H^i(X, F_\star \otimes \omega _X(D)) \simeq H^{n-i}( F_\star ^\vee \otimes \widehat{O_X})^\vee \simeq H^{n-i}(X, \widehat{F_\star ^\vee })^\vee \) .
Now, taking \(E_\star \)  as in the statement to be
\(F_\star \otimes \omega _X(D)\) , we find \(F_\star ^\vee = E_\star ^\vee \otimes \omega _X(D)\)  and so
\(H^i(X, E_\star ) \simeq H^{n-i}(X, {}{{{*[{E_\star ^\vee \otimes \omega _X(D)}]{\hspace{-0.6pt}\bigwedge \hspace{-0.6pt}}{\rule [-/2]{1ex}{}}}{}}{0.5ex}}[1pt]{E_\star ^\vee \otimes \omega _X(D)}{})^\vee \simeq H^{n-i}(X, \widehat{E_\star ^\vee } \otimes \omega _X(D))^\vee .\)
